Air is filled inside a jar which has a pressure gauge connected to it. The temperature of the air inside the jar is same as outside temperature `(= T_(0))` but pressure `(P_(1))` is slightly larger than the atmospheric pressure `(P_(0))`. The stopcock is quickly opened and quickly closed, so that the pressure inside the jar becomes equal to the atmospheric pressure `P_(0)`. The jar is now allowed to slowly warm up to its original temperature `T_(0)`. At this time the pressure of the air inside is `P_(2) (P_(0) lt P_(2) lt P_(1))`. Assume air to be an ideal gas. Calculate the ratio of specific heats `(= gamma)` for the air, in terms of `P_(0), P_(1)` and `P_(2)`.

Answer

`gamma = (ln(P_(0)//P_(1)))/(ln(P_(2)//P_(1)))`